Question

初一数学题求解。。。。。~~

Answer 1

原式
=(2^16)×2^15×(0.2÷0.2)^2010
=2^31×1
=2^32

(2x-3y+3)(2x-3y-7)+25
=(2x-3y)²-7(2x-3y)+3(2x-3y)-21+25
=(2x-3y)²-4(2x-3y)+4
=[(2x-3y)-2]²
=(2x-3y-2)²

M=(-2x²y^4+3/2x)×(5/3x²y)
=-10/3x^4y^5+5/2x³y

(x-y)³+4x²y-4xy²
=(x-y)³+4xy(x-y)
=(x-y)[(x-y)²+4xy]
=(x-y)(x²+2xy+y²)
=(x-y)(x+y)²

(m²+3m)²-8(m²+3m)-20
=[(m²+3m)-10][(m²+3m)+2]
=(m²+3m-10)(m²+3m+2)
=(m+5)(m-2)(m+1)(m+2)

a²bc-3a²c²+2abc-6ac²
=a²c(b-3c)+2ac(b-3c)
=(a²c+2ac)(b-3c)
=ac(a+2)(b-3c)